Question

Solve the following equations:
$2{\tan }^{-1}\left(\cos x\right)={\tan }^{-1}\left(2\csc x\right)$

Answer 1

We know that

$2{\tan }^{-1}x={\tan }^{-1}\frac{2x}{1-x^2}$

Replacing $x$ with $\cos x$

$2{\tan }^{-1}\cos x={\tan }^{-1}\frac{2\cos x}{1-{\cos }^{2}x}$

Using ${\cos }^{2}x+{\sin }^2=1$

i.e. ${\sin }^{2}x=1-{\cos }^{2}x$

$\therefore 2{\tan }^{-1}\cos x={\tan }^{-1}\frac{2\cos x}{{\sin }^{2}x}....\left(1\right)$

Given

$2{\tan }^{-1}\left(\cos x\right)={\tan }^{-1}\left(2\csc x\right)$

Putting values

${\tan }^{-1}\frac{2\cos x}{{\sin }^{2}x}={\tan }^{-1}\left(2\csc x\right)$

$\frac{2\cos x}{{\sin }^{2}x}=2\csc x$

$\frac{\cos x}{{\sin }^{2}x}=\csc x$

$\frac{\cos x}{{\sin }^{2}x}=\frac{1}{\sin x}$

$\cos x=\frac{{\sin }^{2}x}{\sin x}$

$\cos x=\sin x$

$\frac{\sin x}{\cos x}=1$

$\tan x=1$

$\tan x=\tan \frac{\pi }{4}$

$x=\frac{\pi }{4}$